Optimal. Leaf size=207 \[ \frac{(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(5 A-2 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(115 A-43 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
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Rubi [A] time = 0.714781, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2978, 2984, 2985, 2649, 206, 2773} \[ \frac{(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt{a \cos (c+d x)+a}}-\frac{(5 A-2 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(115 A-43 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a \cos (c+d x)+a)^{3/2}}-\frac{(A-B) \tan (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2978
Rule 2984
Rule 2985
Rule 2649
Rule 206
Rule 2773
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac{\int \frac{\left (a (5 A-B)-\frac{5}{2} a (A-B) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{\int \frac{\left (\frac{1}{2} a^2 (35 A-11 B)-\frac{3}{4} a^2 (15 A-7 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{\int \frac{\left (-4 a^3 (5 A-2 B)+\frac{1}{4} a^3 (35 A-11 B) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}+\frac{(115 A-43 B) \int \frac{1}{\sqrt{a+a \cos (c+d x)}} \, dx}{32 a^2}-\frac{(5 A-2 B) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx}{2 a^3}\\ &=-\frac{(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}-\frac{(115 A-43 B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{16 a^2 d}+\frac{(5 A-2 B) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^2 d}\\ &=-\frac{(5 A-2 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac{(115 A-43 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \cos (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B) \tan (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac{(15 A-7 B) \tan (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac{(35 A-11 B) \tan (c+d x)}{16 a^2 d \sqrt{a+a \cos (c+d x)}}\\ \end{align*}
Mathematica [A] time = 3.0522, size = 142, normalized size = 0.69 \[ \frac{\tan (c+d x) (10 (11 A-3 B) \cos (c+d x)+(35 A-11 B) \cos (2 (c+d x))+67 A-11 B)+8 (115 A-43 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-128 \sqrt{2} (5 A-2 B) \cos ^5\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )}{32 d (a (\cos (c+d x)+1))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 5.265, size = 1122, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.46193, size = 1067, normalized size = 5.15 \begin{align*} -\frac{\sqrt{2}{\left ({\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (115 \, A - 43 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} + 2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 16 \,{\left ({\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2} +{\left (5 \, A - 2 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \,{\left ({\left (35 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (11 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 16 \, A\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 3.76784, size = 552, normalized size = 2.67 \begin{align*} \frac{2 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{2 \, \sqrt{2}{\left (A a^{5} - B a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8}} + \frac{\sqrt{2}{\left (21 \, A a^{5} - 13 \, B a^{5}\right )}}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{\sqrt{2}{\left (115 \, A \sqrt{a} - 43 \, B \sqrt{a}\right )} \log \left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a^{3}} - \frac{32 \,{\left (5 \, A \sqrt{a} - 2 \, B \sqrt{a}\right )} \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{a^{3}} + \frac{32 \,{\left (5 \, A \sqrt{a} - 2 \, B \sqrt{a}\right )} \log \left ({\left |{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{a^{3}} + \frac{128 \, \sqrt{2}{\left (3 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt{a} - A a^{\frac{3}{2}}\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} a^{2}}}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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